20,mL of 0.1,M,NH_{4}OH is mixed with 40,mL o | vedclass

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(C) Step $1$: Calculate millimoles of reactants.$n(NH_{4}OH) = 20\,mL 	imes 0.1\,M = 2\,mmol$.$n(HCl) = 40\,mL 	imes 0.05\,M = 2\,mmol$.Step $2$: Re

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$5$

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(C) Step $1$: Calculate millimoles of reactants.$n(NH_{4}OH) = 20\,mL 	imes 0.1\,M = 2\,mmol$.$n(HCl) = 40\,mL 	imes 0.05\,M = 2\,mmol$.Step $2$: Reaction stoichiometry.$NH_{4}OH + HCl ightarrow NH_{4}Cl + H_{2}O$.Since both reactants are $2\,mmol$,they react completely to form $2\,mmol$ of $NH_{4}Cl$ (a salt of weak base and strong acid).Step $3$: Calculate concentration of salt.Total volume $= 20 + 40 = 60\,mL$.$[NH_{4}Cl] = C = \frac{2\,mmol}{60\,mL} = \frac{1}{30}\,M$.Step $4$: Calculate $pH$ of salt solution.$pH = \frac{1}{2}[pK_{w} - pK_{b} - \log C]$.$pK_{w} = 14$,$pK_{b} = -\log(10^{-5}) = 5$.$pH = \frac{1}{2}[14 - 5 - \log(1/30)] = \frac{1}{2}[9 + \log 30] = \frac{1}{2}[9 + \log 3 + \log 10] = \frac{1}{2}[9 + 0.48 + 1] = \frac{10.48}{2} = 5.24$.The nearest value is $5.2$.

$20\,mL$ of $0.1\,M\,NH_{4}OH$ is mixed with $40\,mL$ of $0.05\,M\,HCl$. The $pH$ of the mixture is nearest to. (Given: $K_{b}(NH_{4}OH) = 1 \times 10^{-5}, \log 2 = 0.30, \log 3 = 0.48, \log 5 = 0.69, \log 7 = 0.84, \log 11 = 1.04$) (in $.2$)

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